how to replace space with %20 in url in android in jsonparsing

Question

In my application i want to replace space with %20 in my string.I tried in this way,

String flag1="http://74.208.194.142/admin/upload_image/1342594079_images (2).jpg";
flag1.replaceAll("", "%20");

But it is not working please help me.I am getting null pointer exception.

This is not android. This is java. Very basic java. – keyser Oct 01 '12 at 10:24 — keyser, Oct 01 '12 at 10:24

score 10 · Answer 1 · answered Oct 01 '12 at 10:23

10

You should do it like:

flag1 = flag1.replaceAll(" ", "%20");

first you were putting empty string instead of space.. and second you must return the value into the flag1 variable..

answered Oct 01 '12 at 10:23

Nermeen

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please help me where i did mistake suggest me where i need to change – user1083266 Oct 01 '12 at 10:31
flag1 = flag1.replaceAll(" ", "%20"); should work – Nermeen Oct 01 '12 at 10:33
check also http://stackoverflow.com/a/2593319/1434631 – Nermeen Oct 01 '12 at 10:35
i tried that also but i am getting null pointer exception – user1083266 Oct 01 '12 at 10:36
1

What did you try? The only way that `flag1 = flag1.replaceAll(" ", "%20");` can give an NPE is if `flag1` was null to start with. – Stephen C Oct 01 '12 at 10:54

thobens · Answer 2 · 2012-10-01T12:32:25.990

5

Have a look at http://docs.oracle.com/javase/1.5.0/docs/api/java/net/URLEncoder.html

It is not only the blanks that need to be replaced. URLs consist of special caracters for special meanings (e.g. the question mark to start the query string)

String flag1 = URLEncoder.encode("This string has spaces", "UTF-8")

edited Oct 01 '12 at 12:32

answered Oct 01 '12 at 12:19

thobens

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6

Not sure why this is even voted up, OP is replacing path arguments, not query params, URLEncoder encodes " " to "+" not "%20" – Edison May 02 '14 at 17:41
This will add + not %20 – DjHacktorReborn Nov 14 '15 at 18:38

score 4 · Answer 3 · answered Oct 01 '12 at 10:25

4

In flag1.replaceAll(), your space is missing. try:

String flag1="http://74.208.194.142/admin/upload_image/1342594079_images (2).jpg";
flag1 = flag1.replaceAll(" ", "%20");

and set the result to the flag1.

answered Oct 01 '12 at 10:25

stealthjong

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score 4 · Answer 4 · answered Oct 01 '12 at 10:38

    String flag1="http://74.208.194.142/admin/upload_image/1342594079_images (2).jpg";
    URI uri = null;
    try {
        uri = new URI(flag1.replaceAll(" ", "%20"));
    } catch (URISyntaxException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    System.out.println(uri);

The console will display output as

http://74.208.194.142/admin/upload_image/1342594079_images%20(2).jpg

how to replace space with %20 in url in android in jsonparsing

4 Answers4

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